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My dad had exactly that same experience on a mid 50's model Ford. His
thermostat failed closed, so he took it out and didn't replace it.
The car overheated. His analysis was the obvious one that the water
went thru the radiator too fast to cool. It was years before I ever
found out what was really happening, and AFAIK he never did.
Bob White
On Fri, 13 Aug 2004 15:31:51 -0400
"ericruttan@chartermi.net" <ericruttan@chartermi.net> wrote:
> Let me add a trifle.
>
> Car racers I ran with as a kid with less creativity and more "results
> oriented" said the same thing, not because they had any fancy
> thermometers, but because they raced with the thermostats out, and
> boiled over.
>
> The reasoning was hard to argue with. Put the thermostat (read
> restriction) in and the car ran fine. Take it out and it would boil
> over.
>
> How can a student of physics argue with that!
>
> Years later I found out that the water pump was cavitating due to the
> reduced pressure. With the restriction the pressure was higher and
> the cavitating less likely.
>
> With the addition of a better water pump vane design the boil overs
> stopped when the thermostats where removed, and the cars ran cooler,
> and faster.
>
>
> Ed Anderson wrote:
> > Ok, David, since you asked.
> >
> > In the early days of car racing (and still) some folks decided to do
> > some data collection about the effects of different factors.
> >
> > Being aware of the value of data they placed "Mom's" cooking
> > thermometers at the entrance and exit of a radiator mounted next to
> > the still. With a fire burning under the still they proceeded to
> > pump hot water .... OK, Ok, getting serious.
> >
> > Here is what let to the mistaken belief that "Slow Water cools
> > better".
> >
> > Folks observed that if they measure the entry and exit temperature
> > of water as it passed through a radiator an interesting thing. The
> > slower the water flowed the greater the temperature difference
> > between the water entering and the water exiting the radiator. AND
> > that observation is absolutely correct. The reason of course is the
> > longer the water stays in the radiator the more heat is removed from
> > the water before it exits the radiator. So far they were on solid
> > ground in their observations - however their interpretation of the
> > data is where they went afoul.
> >
> > The assumed the greater temperature drop of the slower water was
> > good as it showed more heat went out of the water - I mean how can
> > you argue with that?? Therefore slow water must be better at
> > cooling - right? Well, actually NOT!!!
> >
> > Where they failed is in considering the effects of the "slower"
> > water on the ENTIRE system - not just the radiator, but the engine
> > which this is all about in the first place. In reality lessening
> > the flow of the coolant(while it will cause a great deltaT of the
> > coolant flowing through the radiator) will fail to remove heat from
> > the engine as fast as faster moving water will as the recent
> > examples show. However, there are folks to this day will swear by
> > "Slow" water. I argued with old man Lou Ross for 45 minutes on the
> > phone one time trying to reason it through with him but to no avail.
> > It only stopped when I got a bit
> > frustrated and stated "Well, Lou, if slow water cools better ---
> > then Stopped water must cool best!!"" He of course knew that wasn't
> > the case, but still clung to that belief.
> >
> > Now, I have read that in some cases restrictors have helped cooling
> > - but not by slowing the water flow. In some cases, it supposedly
> > reduced cavitation of the water pump, kept head pressure in the
> > block higher, promoted nucleated boiling and a number of things that
> > I never bothered to look into. But all things equal more coolant
> > flow equals more heat removed from the engine (always assuming you
> > get rid of the heat through a radiator before the coolant returns to
> > the engine).
> >
> > There David, hope that answers your question.
> >
> > Ed
> >
> > Ed Anderson
> > RV-6A N494BW Rotary Powered
> > Matthews, NC
> > ----- Original Message -----
> > From: "David Carter" <dcarter@datarecall.net>
> > To: "Rotary motors in aircraft" <flyrotary@lancaironline.net>
> > Sent: Friday, August 13, 2004 11:38 AM
> > Subject: [FlyRotary] Re: Answer to when is 2 gallons enough?Re:
> > DeltaT Coolant was : [FlyRotary] Re: coolant temps
> >
> >
> >
> >>Ed, I really like your explanations - the math and attention to
> >explaining>the units. Good work.
> >>
> >>Now, about "slower water cooling better". I'd like to hear "the
> >rest of
> >
> > the
> >
> >>story". Here's why I ask: I went into a car "window tint" shop 3
> >weeks>ago to shop for tint on 3 windows on south side of my house.
> >Took care of>that busines - and noticed some nice after-market
> >anodized blue aluminum>housings and stuff hanging on a wall display.
> >I asked if this was an>electric water pump and asked a question about
> >it. The "tint" guy said,>"You'll have to ask ____, the speed shop
> >guy, who shares the 4 shop bays>with me."
> >>
> >>I went out and visited with the racing guy - yep, used electric
> >water
> >
> > pumps
> >
> >>and can fix me up with an in-line pump for my rotary engine on RV-6
> >since>aluminum housings are only for specific V-8s. He was working
> >on his drag>racer and pointed to the thermosat housing and some large
> >washers. "Then>thar washers are different sizes for restricting and
> >adjusting the flow
> >
> > rate
> >
> >>through the radiator. Makes it cool better." So, "thar you have
> >it".>Doesn't make sense to me
> >>
> >>David
> >>
> >>----- Original Message -----
> >>From: "Ed Anderson" <eanderson@carolina.rr.com>
> >>To: "Rotary motors in aircraft" <flyrotary@lancaironline.net>
> >>Sent: Friday, August 13, 2004 9:14 AM
> >>Subject: [FlyRotary] Answer to when is 2 gallons enough?Re: DeltaT
> >Coolant>was : [FlyRotary] Re: coolant temps
> >>
> >>
> >>I'm sorry, Mark, I did not show that step. You are correct the
> >weight>(mass) of water(or any other cooling medium) is an important
> >factor as is>its specific heat.
> >>
> >> In the example you used - where we have a static 2 gallons
> >capacity of>water, It would actually only take 8*2 = 16 lbms *10 =
> >160 BTU to raise
> >
> > the
> >
> >>temp of the water 1 degree F. The difference is in one case we are
> >
> > talking
> >
> >>about raising the temperature of a fixed static amount of water
> >which can>not readily get rid of the heat, in the other (our radiator
> >engine case)
> >
> > we
> >
> >>are talking about how much heat the coolant can transfer from engine
> >to>radiator. Here the flow rate is the key factor.
> >>
> >>But lets take your typical 2 gallon cooling system capacity and see
> >what
> >
> > we
> >
> >>can determine.
> >>
> >>If we take our 2 gallons and start moving it from engine to radiator
> >and>back we find that each times the 2 gallons circulates it
> >transfers 160 BTU>(in our specific example!!). So at our flow rate of
> >30 gpm we find it will>move that 160 BTU 15 times/minute (at 30
> >gallons/minute the 2 gallons
> >
> > would
> >
> >>be transferred 15 times). So taking our 160 BTU that it took to
> >raise the>temp of our 2 gallons of static water 10F that we now have
> >being moved
> >
> > from
> >
> >>engine to radiator 15 times a minute = 160*15 = 2400 BTU/Min.
> >Amazing
> >
> > isn't
> >
> >>it? So no magic, just math {:>). So that is how our 2 gallons of
> >water>can transfer 2400 BTU/min from engine to radiator. It also
> >shows why the>old wives tale about "slow water" cooling better is
> >just that (another
> >
> > story
> >
> >>about how that got started)
> >>
> >>
> >>In the equation Q = W*deltaT*cp that specifies how much heat is
> >
> > transferred
> >
> >>,we are not talking about capacity such as 2 gallons capacity of a
> >cooling>system but instead are talking about mass flow. As long as
> >we reach that>flow rate 1 gallon at 30 gpm or 1/ gallon at 60 gpm or
> >1/4 gallon at 120>gpm all will remove the same amount of heat.
> >However if you keep
> >
> > increasing
> >
> >>the flow rate and reducing the volume you can run into other
> >problems -
> >
> > like
> >
> >>simply not enough water to keep your coolant galleys filled {:>), so
> >there>are limits.
> >>
> >>Our 2 gallon capacity is, of course, simply recirculated at the
> >rate of
> >
> > 30
> >
> >>gpm through our engine (picking up heat- approx 2400 BTU/min in this
> >>specific example) and then through our radiator (giving up heat of
> >2400>BTU/Min to the air flow through the radiators) assuming
> >everything works
> >
> > as
> >
> >>planned. IF the coolant does not give up as much heat in the
> >radiators
> >
> > (to
> >
> >>the air stream) as it picks up in the engine then you will
> >eventually>(actually quite quickly) over heat your engine.
> >>
> >>The 240 lb figure I used in the previous example comes from using 8
> >lb/gal>(a common approximation, but not precise as you point out) to
> >calculate
> >
> > the
> >
> >>mass flow.
> >>
> >>The mass flow = mass of the medium (8 lbs/gallon for water) * Flow
> >rate(30>gpm) =240 lbs/min mass flow. Looking at the units we have
> >>(8 lbs/gallon)*(30 Gallon/minute) canceling out the like units
> >(gallons)>leaves us with 240 lb/minute which is our mass flow in this
> >case.>
> >>Then using the definition of the BTU we have 240 lbs of water that
> >must be>raised 10F. Using our heat transfer equation
> >>
> >>Q = W*deltaT*cp, we have Q = 240*10*1 = 2400 BTU/minute is required
> >to>increase the temperature of this mass flow by 10F
> >>
> >>Using the more accurate weight of water we would have 8.34*30 =
> >250.2>lbm/minute so the actual BTU required is closer to 2502
> >BTU/min instead
> >
> > of
> >
> >>my original 2400 BTU/Min, so there is apporx a 4% error in using 8
> >>lbs/gallon. If we could ever get accurate enough where this 4% was
> >an>appreciable part of the total errors in doing our back of the
> >envelope>thermodynamics then it would pay to use 8.34 vice 8, but I
> >don't think we>are there, yet {:>).
> >>
> >>Now the same basic equation applies to the amount of heat that the
> >air>transfers away from out radiators. But here the mass of air is
> >much lower>than the mass of water so therefore it takes a much higher
> >flow rate to>equal the same mass flow. What makes it even worse is
> >that the specific>heat of air is only 0.25 compared to water's 1.0.
> >So a lb of air will
> >
> > only
> >
> >>carry approx 25% the heat of a lb of water, so again for this reason
> >you>need more air flow.
> >>
> >>if 30 gpm of water will transfer 2400 bTu of engine heat (using
> >Tracy's
> >
> > fuel
> >
> >>burn of 7 gallon/hour), how much air does it require to remove that
> >heat>from the radiators?
> >>
> >>Well again we turn to our equation and with a little algebra we
> >have W =>Q/(DeltaT*Cp) = 2400/(10*1) = 240 lbm/min. Not a surprise as
> >that is what
> >
> > we
> >
> >>started with.
> >>
> >>But now taking the 240 lbm/min mass flow and translating that into
> >Cubic>feet/minute of air flow. We know that a cubic foot of air at
> >sea level>weighs approx 0.076 lbs. So 240 lbm/(0.076 lbm/Cubic foot)
> >= 3157 cubic>feet/min to equal the same mass as the coolant. But
> >since the specific
> >
> > heat
> >
> >>of air is lower (0.25) that water, we actually need 75% more air
> >mass or>1.75 * 3157 = 5524.75 CFM air flow at sea level. Now I know
> >this sounds
> >
> > like
> >
> >>a tremendous amount of air but stay with me through the next step.
> >>
> >>Taking two GM evaporator cores with a total frontal area of 2*95 =
> >190 sq>inches and turning that in to square feet = 1.32 sq ft we take
> >our>5524.75 cubic feet minute and divide by 1.32 sq ft = 4185 ft/min
> >for the>required air velocity to move that much air volume through
> >our two>evaporator cores. To get the air velocity in ft/sec divide
> >4185/60 =
> >
> > 69.75
> >
> >>ft/sec airflow velocity through our radiators or 47.56 Mph. Now
> >that>sounds more reasonable doesn't it??
> >>
> >>Now all of this is simply a first order estimate. There are lots of
> >
> > factors
> >
> >>such as the density of the air which unlike water changes with
> >altitude,
> >
> > the
> >
> >>temperature of the air, etc. that can change the numbers a bit.
> >But, then>there is really not much point in trying to be more
> >accurate given the>limitations of our experimentation accuracy {:>).
> >>
> >>
> >>Also do not confuse the BTUs required to raise the temperature of 1
> >lb of>water 1 degree F with that required to turn water in to vapor -
> >that>requires orders of magnitude more BTU.
> >>
> >>Hope this helped clarify the matter.
> >>
> >>Ed
> >>
> >>
> >>Ed Anderson
> >>RV-6A N494BW Rotary Powered
> >>Matthews, NC
> >> ----- Original Message -----
> >> From: Mark Steitle
> >> To: Rotary motors in aircraft
> >> Sent: Friday, August 13, 2004 8:32 AM
> >> Subject: [FlyRotary] Re: DeltaT Coolant was : [FlyRotary] Re:
> >coolant>temps
> >>
> >>
> >> Ed,
> >> Please humor me (a non-engineer) while I ask a dumb question. If
> >it
> >
> > takes
> >
> >>1BTU to raise 1lb of water 1 degree, and you factor in 30 gpm flow
> >to come>up with a 2400 BTU requirement for a 10 degree rise for 1 lb
> >of water,
> >
> > where
> >
> >>does the number of pounds of water figure into the equation, or do
> >we just>ignore that issue? Water is 8.34 lbs/gal, and say you have 2
> >gallons of>coolant, that would be 16.68 lbs. Seems that we would
> >need to multiply
> >
> > the
> >
> >>2400 figure by 16.68 to arrive at a total system requirement of
> >40,032>BTU/min. What am I missing here?
> >>
> >> Mark S.
> >>
> >>
> >> At 09:58 PM 8/12/2004 -0400, you wrote:
> >>
> >> Right you are, Dave
> >>
> >> Below is one semi-official definition of BTU in English units.
> >1 BTU>is amount of heat to raise 1 lb of water 1 degree Fahrenheit.
> >>
> >> So with Tracy's 30 gpm flow of water = 240 lbs/min. Since its
> >>temperature is raised 10 degree F we have
> >>
> >> BTU = 240 * 10 * 1 = 2400 BTU/min
> >>
> >> I know I'm ancient and I should move into the new metric world,
> >but
> >
> > at
> >
> >>least I didn't do it in Stones and Furlongs {:>)
> >>
> >> Ed
> >>
> >> The Columbia Encyclopedia, Sixth Edition. 2001.
> >>
> >> British thermal unit
> >>
> >>
> >> abbr. Btu, unit for measuring heat quantity in the customary
> >system of>English units of measurement, equal to the amount of heat
> >required to
> >
> > raise
> >
> >>the temperature of one pound of water at its maximum density [which
> >occurs>at a temperature of 39.1 degrees Fahrenheit (°F) ] by 1°F. The
> >Btu may
> >
> > also
> >
> >>be defined for the temperature difference between 59°F and 60°F. One
> >Btu
> >
> > is
> >
> >>approximately equivalent to the following: 251.9 calories; 778.26
> >>foot-pounds; 1055 joules; 107.5 kilogram-meters; 0.0002928
> >kilowatt-hours.
> >
> > A
> >
> >>pound (0.454 kilogram) of good coal when burned should yield 14,000
> >to>15,000 Btu; a pound of gasoline or other .
> >>
> >>
> >>
> >>
> >>
> >>
> >>
> >> Ed Anderson
> >> RV-6A N494BW Rotary Powered
> >> Matthews, NC
> >> ----- Original Message -----
> >> From: DaveLeonard
> >> To: Rotary motors in aircraft
> >> Sent: Thursday, August 12, 2004 8:12 PM
> >> Subject: [FlyRotary] Re: DeltaT Coolant was : [FlyRotary] Re:
> >
> > coolant
> >
> >>temps
> >>
> >>
> >> Ed, are those units right. I know that the specific heat of
> >water
> >
> > is
> >
> >>1.0 cal/(deg Celsius*gram). Does that also work out to 1.0
> >BTU/(deg.>Farhengight * Lb.) ?
> >>
> >> Dave Leonard
> >> Tracy my calculations shows your coolant temp drop is where it
> >
> > should
> >
> >>be:
> >>
> >> My calculations show that at 7 gph fuel burn you need to get
> >rid of>2369 BTU/Min through your coolant/radiators. I rounded it off
> >to 2400>BTU/min.
> >>
> >> Q = W*DeltaT*Cp Basic Heat/Mass Flow equation With water as
> >the
> >
> > mass
> >
> >>with a weight of 8 lbs/ gallon and a specific heat of 1.0
> >>
> >> Q = BTU/min of heat removed by coolant mass flow
> >>
> >> Assuming 30 GPM coolant flow = 30*8 = 240 lb/min mass flow.
> >
> > specific
> >
> >>heat of water Cp = 1.0
> >>
> >>
> >> Solving for DeltaT = Q/(W*Cp) = 2400/(240*1) = 2400/240 =
> >10 or>your delta T for the parameters specified should be around 10F
> >>
> >> Assuming a 50/50 coolant mix with a Cp of 0.7 you would have
> >approx>2400/(240 *0.7) = 2400/168 = 14.2F so I would say you do not
> >fly with>
> >> a 50/50 coolant mix but something closer to pure water. But
> >in any>case, certainly in the ball park.
> >>
> >> You reported 10-12F under those conditions, so I would say
> >condition>is 4. Normal operation
> >>
> >> Ed
> >>
> >> Ed Anderson
> >> RV-6A N494BW Rotary Powered
> >> Matthews, NC
> >>
> >>
> >>
> >>
> >>>> Homepage: http://www.flyrotary.com/
> >>>> Archive: http://lancaironline.net/lists/flyrotary/List.html
> >
> >
> >
> >
> >>> Homepage: http://www.flyrotary.com/
> >>> Archive: http://lancaironline.net/lists/flyrotary/List.html
> >
> >
>
>
> >> Homepage: http://www.flyrotary.com/
> >> Archive: http://lancaironline.net/lists/flyrotary/List.html
>
>
--
http://www.bob-white.com
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